Evaluate the indefinite integral: $$\int\sqrt{3-t}\;dt$$

Evaluate the indefinite integral: $$\int x^2\sqrt{x^3+9}\;dx, \quad u = x^3+9$$

I am confused on how to evaluate the two problems above. On first problem I've got:

u^2=sqrt(3-t) = u=3-t == t=3/u

But from there i'm lost. On the second problem ive got du=3x^2dx, but I'm confused what to do with the du. I missed a class due to sickness, tried to figure it out through some of the stuff I've found online but can't seem to put the pieces together. Any and all help is much appreciated.

  • 1
    $\begingroup$ Foe the first problem, the proposed substitution is not best. Quite natural is $u=3-t$, which is not the same as $u^2=\sqrt{3-t}$. From $u=3-t$ we get $du=-dt$, or equivalently $dt=-du$. So we want $\int -u^{1/2}\,du$. You probably know how to integrate powers. We get $-\frac{2}{3}u^{3/2}+C$, that is, $-\frac{2}{3}(3-t)^{3/2}+C$. $\endgroup$ – André Nicolas Apr 7 '14 at 1:59

First, let $u = 3-t$. Then, $du = \frac{\partial u}{\partial t} dt = - dt$. Hence, the integrand is $- \sqrt{u} du$. This integrates to $-2/3\cdot (3-t)^{3/2}+c$. Why? Since the integral of $x^{n}$ is $\frac{x^{n+1}}{n+1}+c$.

The second problem is similar. Let $u = x^3 + 9$. You are integrating $1/3\cdot du\cdot \sqrt{u}$, since $du = \frac{\partial u}{\partial x} dx = 3x^2 dx$. This evaluates to $2/9\cdot (x^3+9)^{3/2}+c$.

  • $\begingroup$ @user102817, does this help? Or would like more steps? $\endgroup$ – Chris K Apr 7 '14 at 21:54

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